document.write( "Question 1053378: A bag containing 5 white, and 7 black balls. All of the balls are drawn at random from the bag one by one without replacement. Find the probability of drawing white ball on 6th attempt and the previous 3 balls being black. \n" ); document.write( "

Algebra.Com's Answer #668623 by Edwin McCravy(20060) $\"\"$ $\"About$

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A bag containing 5 white, and 7 black balls. All of the balls are drawn at
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document.write( "We want to know the probability of drawing them in this order,\r\n" );
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document.write( "X,X,B,B,B,W,X,X,X,X,X,X\r\n" );
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document.write( "where the X's can be either color.\r\n" );
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document.write( "Since it doesn't matter what is drawn after the 6th ball, we only need\r\n" );
document.write( "to consider the probability that the first 6 drawings are in this\r\n" );
document.write( "order:\r\n" );
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document.write( "X,X,B,B,B,W\r\n" );
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document.write( "So there are four cases for the first 6 balls drawn:\r\n" );
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document.write( "Case 1: W,W,B,B,B,W\r\n" );
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document.write( "Probability = (5/12)(4/11)(7/10)(6/9)(5/8)(3/7) = 5/264\r\n" );
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document.write( "Case 2: W,B,B,B,B,W\r\n" );
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document.write( "Probability = (5/12)(7/11)(6/10)(5/9)(4/8)(4/7) = 5/198\r\n" );
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document.write( "Case 3: B,W,B,B,B,W\r\n" );
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document.write( "Probability = (7/12)(5/11)(6/10)(5/9)(4/8)(4/7) = 5/198\r\n" );
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document.write( "Case 4: B,B,B,B,B,W \r\n" );
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document.write( "Probability = (7/12)(6/11)(5/10)(4/9)(3/8)(5/7) = 5/264\r\n" );
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document.write( "The probability that one of those cases will occur equals\r\n" );
document.write( "the sum of those probabilities, which is:\r\n" );
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document.write( "5/264 + 5/198 + 5/198 + 5/264 = 35/396\r\n" );
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document.write( "Edwin

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